Write a function, persistence, that takes in a positive parameter n and returns its multiplicative persistence, which is the number of times you must multiply the digits in n until you reach a single digit.
For example:
I have tried this code which works but the game's rule is that I cannot pass 2 arguments. So I have to eliminate the counter argument.
def persistence(n,counter): #recursion
n = list(str(n))
product = 1
for i in n:
product *= int(i)
if product < 10:
return counter+1
else:
counter +=1
return persistence(product,counter)
persistence(39) # returns 3, because 3*9=27, 2*7=14, 1*4=4
# and 4 has only one digit
You could use a default argument if you wanted to keep your algorithm the same:
def persistence(n, counter=0):
product = 1
for i in str(n):
product *= int(i)
if product < 10:
return counter + 1
return persistence(product, counter + 1)
print(persistence(39))
Having said that, as @TomKarzes points out, there's no need for the counter argument at all (or recursion for that matter)--pass the result back up the call stack rather than down. Any given call doesn't need any information from previous calls other than the current state of n itself to determine persistence.
def persistence(n):
product = 1
for i in str(n):
product *= int(i)
if product < 10:
return 1
return persistence(product) + 1
print(persistence(39))